Code-EE

Test Booklet Code

DATE : 06/05/2018

EE CHLAA Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.: 011-47623456 Fax : 011-47623472

Time : 3 hrs.

Answers & Solutions

Max. Marks : 720

for NEET (UG) - 2018 Important Instructions : 1.

The test is of 3 hours duration and Test Booklet contains 180 questions. Each question carries 4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response, one mark will be deducted from the total scores. The maximum marks are 720.

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The CODE for this Booklet is EE.

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1

NEET (UG) - 2018 (Code-EE) CHLAA

1.

At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere?

3.

(Given :

(1) 26.8%

Mass of oxygen molecule (m) = 2.76 × 10–26 kg

(2) 6.25%

Boltzmann's constant kB = 1.38 × 10–23 JK–1)

(3) 20%

(1) 2.508 ×

104

K

(2) 5.016 ×

(3) 8.360 × 104 K

104

K

(4) 12.5%

(4) 1.254 × 104 K

Answer ( 1 )

 T  S o l . Efficiency of ideal heat engine,    1 2  T1   T2 : Sink temperature

Answer ( 3 ) S o l . Vescape = 11200 m/s Say at temperature T it attains Vescape

So,

T1 : Source temperature

3kB T  11200 m/s mO2

T   %   1  2   100 T1  

On solving,

273     1   100 373  

4

T = 8.360 × 10 K 2.

The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is

The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

 100     100  26.8%  373 

4.

The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is (1) 13.2 cm (2) 12.5 cm (3) 8 cm

2 (1) 5

(3)

1 (2) 3

2 3

(4)

(4) 16 cm Answer ( 1 )

2 7

S o l . For closed organ pipe, third harmonic 

Answer ( 1 )

For open organ pipe, fundamental frequency

S o l . Given process is isobaric

dQ  n Cp dT



v 2l 

Given,

5  dQ  n  R  dT 2 

3v v  4l 2l 

dW  P dV = n RdT

Required ratio 

3v 4l

 l 

dW nRdT 2   dQ 5 5  n  R  dT 2 



2

4l 2l  32 3 2  20  13.33 cm 3


5.

A metallic rod of mass per unit length 0.5 kg m–1 is lying horizontally on a smooth inclined plane which makes an angle of 30° with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

So, resistance of galvanometer RG 

8.

(1) 7.14 A

(2) 14.76 A

A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from

(3) 5.98 A

(4) 11.32 A

(1) The current source

Answer ( 4 ) S o l . For equilibrium,

B

mg sin30  Il Bcos 30 mg I tan30 lB



6.

IS 51 5000    250  3 VS 20  10 20

s co

(2) The lattice structure of the material of the rod

° 30

(3) The magnetic field (4) The induced electric field due to the changing magnetic field

llB ° 30° llB 30

n si g m 30°

0.5  9.8

Answer ( 1 )

 11.32 A 0.25  3 An inductor 20 mH, a capacitor 100 F and a resistor 50  are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is (1) 0.79 W

(2) 2.74 W

(3) 0.43 W

(4) 1.13 W

S o l . Energy of current source will be converted into potential energy of the rod. 9.

In the circuit shown in the figure, the input voltage Vi is 20 V, VBE = 0 and VCE = 0. The values of IB, IC and  are given by

20 V

Answer ( 1 ) S o l . Pav

RB

Vi

2

V    RMS  R  Z 

RC 4 k C

500 k B

E

2

1   Z  R2   L   56  C  

(1) IB = 40 A, IC = 10 mA,  = 250

2

7.

(2) IB = 20 A, IC = 5 mA,  = 250

  10   50  0.79 W Pav    2 56    Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is



 

(1) 40 

(2) 250 

(3) 25 

(4) 500 

(3) IB = 25 A, IC = 5 mA,  = 200 (4) IB = 40 A, IC = 5 mA,  = 125 Answer ( 4 ) S o l . VBE = 0 VCE = 0 Vb = 0

20 V

Answer ( 2 )

IC

S o l . Current sensitivity IS 

NBA C

Vi

Voltage sensitivity VS 

NBA CRG

3

RB Ib

500 k

Vb

RC = 4 k


(20  0) 4  103 IC = 5 × 10–3 = 5 mA

12.

IC 

Vi = VBE + IBRB Vi = 0 + IBRB 20 = IB × 500 × 103 IB 

 10.

Unpolarised light is incident from air on a plane surface of a material of refractive index ''. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation? (1) Reflected light is polarised with its electric vector parallel to the plane of incidence

20  40 A 500  103

 1 (2) i  sin1   

IC 25  103   125 Ib 40  106

In the combination of the following gates the output Y can be written in terms of inputs A and B as

(3) Reflected light is polarised with its electric vector perpendicular to the plane of incidence

A

 1 (4) i  tan1   

B

Y

Answer ( 3 ) S o l . When reflected light rays and refracted rays are perpendicular, reflected light is polarised with electric field vector perpendicular to the plane of incidence.

(1) A  B (2) A  B  A  B (3) A  B  A  B (4) A  B

i

Answer ( 3 ) Sol. A

B

A

AB

B A B



Y

Also, tan i =  (Brewster angle)

AB

13.

Y  (A  B  A  B) 11.

In a p-n junction diode, change in temperature due to heating (1) Affects only reverse resistance (2) Does not affect resistance of p-n junction (3) Affects only forward resistance (4) Affects the overall V - I characteristics of p-n junction

In Young's double slit experiment the separation d between the slits is 2 mm, the wavelength  of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same  and D) the separation between the slits needs to be changed to (1) 1.8 mm

(2) 2.1 mm

(3) 1.9 mm

(4) 1.7 mm

Answer ( 4 )

Answer ( 3 )

S o l . Due to heating, number of electron-hole pairs will increase, so overall resistance of diode will change.

S o l . Angular width 

Due to which forward biasing and reversed biasing both are changed.

0.20  4

 2 mm

 d

…(i)


0.21 

 d

E0

…(ii)

V0 F

0.20 d Dividing we get, 0.21  2 mm

Acceleration of electron a

 d = 1.9 mm 14.

eE0 m

An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of

Velocity after time ‘t’

(1) Small focal length and large diameter

So,  

eE0 ⎛ V  ⎜ V0  m ⎝

(2) Large focal length and large diameter

h  mV

⎞ t⎟ ⎠

h eE ⎞ ⎛ m ⎜ V0  0 t ⎟ m ⎠ ⎝

(3) Large focal length and small diameter



(4) Small focal length and small diameter Answer ( 2 ) S o l . For telescope, angular magnification =

f0 fE



So, focal length of objective lens should be large. Angular resolution =

16.

D should be large. 1.22

So, objective should have large focal length (f0) and large diameter D. 15.

An electron of mass m with an initial velocity

h ⎡ eE0 ⎤ mV0 ⎢1  t⎥ ⎣ mV0 ⎦ 0 ⎡ eE0 ⎢1  ⎣ mV0

⎤ t⎥ ⎦

For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is (1) 20

(2) 30

(3) 10

(4) 15

Answer ( 1 )



V  V0 î (V 0 > 0) enters an electric field

S o l . Number of nuclei remaining = 600 – 450 = 150 n

N ⎛ 1⎞  N0 ⎜⎝ 2 ⎟⎠



E  –E0 î (E0 = constant > 0) at t = 0. If 0 is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is

t

150 ⎛ 1 ⎞ t 1/2  600 ⎜⎝ 2 ⎟⎠

0

(1)

⎛ eE0 ⎞ t⎟ ⎜1 mV0 ⎠ ⎝ (2) 0t

t

2

⎛ 1⎞ ⎛ 1 ⎞ t 1/2 ⎜2⎟  ⎜2⎟ ⎝ ⎠ ⎝ ⎠

t = 2t1/2 = 2 × 10

⎛ eE0 ⎞ t⎟ (3)  0 ⎜ 1  mV0 ⎠ ⎝

= 20 minute 17.

Answer ( 1 )

The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is

S o l . Initial de-Broglie wavelength

(1) 1 : 1

(2) 2 : –1

(3) 1 : –1

(4) 1 : –2

(4) 0

0 

h mV0

Answer ( 3 ) 5


S o l . KE = –(total energy)

–

So, Kinetic energy : total energy = 1 : –1 18.

When the light of frequency 20 (where 0 is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v 1 . When the frequency of the incident radiation is increased to 5 0, the maximum velocity of electrons emitted from the same plate is v2. The ratio of v1 to v2 is (1) 1 : 2

(2) 4 : 1

(3) 1 : 4

(4) 2 : 1



When object is displaced by 20 cm towards mirror. Now, u2 = –20 1 1 1   f v2 u2

1 mv2 2

h(20 )  h0 

h 0 

1 1 1 –  –15 v2 20

1 mv12 2

1 mv12 2

1 1 1  – v2 20 15

…(i)

v2 = –60 cm

1 h(50 )  h0  mv22 2 1 mv22 2 Divide (i) by (ii), 4h0 

So, image shifts away from mirror by = 60 – 24 = 36 cm. 20.

…(ii)

The refractive index of the material of a prism is 2 and the angle of the prism is 30°. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is

1 v12  4 v22

v1 1  v2 2 19.

1 1 1   v1 –15 40

v1 = –24 cm

Answer ( 1 ) S o l . E  W0 

1 1 1  – 15 v1 40

An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be

(1) 60°

(2) 30°

(3) 45°

(4) Zero

Answer ( 3 )

(1) 30 cm away from the mirror

S o l . For retracing its path, light ray should be normally incident on silvered face.

(2) 30 cm towards the mirror (3) 36 cm away from the mirror (4) 36 cm towards the mirror

30°

Answer ( 3 ) Sol.

i

f = 15 cm O

M

40 cm

60° 30°

 2 Applying Snell's law at M, sin i 2  sin30 1

1 1 1   f v1 u 6


 sin i  2 

sin i  21.

1 2

1 2

I

O



V  Viˆ . The instantaneous oscillating electric field of this em wave is along +y axis. Then the direction of oscillating magnetic field of the em wave will be along velocity

n

I

(3)

(4)

O

(2) –y direction

n

O

n

Answer ( 1 ) n  Sol. I   nr r So, I is independent of n and I is constant.  I

(3) +z direction (4) –x direction Answer ( 3 ) 

O

n

I

(1) –z direction



(2)

(1)

i.e. i = 45°

An em wave is propagating in a medium with a

I



Sol. E  B  V 

ˆ  (B)  Viˆ (Ej) 

So, B  Bkˆ

O n 24. A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I. The value of 'n' is (1) 10 (2) 20 (3) 11 (4) 9 Answer ( 1 ) E Sol. I  ...(i) nR  R E 10 I  ...(ii) R R n Dividing (ii) by (i), (n  1)R 10  1   n  1 R   After solving the equation, n = 10 25. A carbon resistor of (47 ± 4.7) k is to be marked with rings of different colours for its identification. The colour code sequence will be (1) Violet – Yellow – Orange – Silver (2) Yellow – Green – Violet – Gold (3) Yellow – Violet – Orange – Silver (4) Green – Orange – Violet – Gold Answer ( 3 )

Direction of propagation is along +z direction. 22.

The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance (1) 0.138 H

(2) 1.389 H

(3) 138.88 H

(4) 13.89 H

Answer ( 4 ) S o l . Energy stored in inductor U

1 2 Ll 2

25  10 –3  L 

1  L  (60  10 –3 )2 2

25  2  106  10–3 3600 500 36

= 13.89 H 23.

A battery consists of a variable number 'n' of identical cells (having internal resistance 'r' each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shows the correct relationship between I and n? 7


S o l . (47 ± 4.7) k = 47 × 103 ± 10%

Sol. h 

 Yellow – Violet – Orange – Silver 26.

A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27ºC two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27ºC is (1) 330 m/s

(2) 350 m/s

(3) 339 m/s

(4) 300 m/s

29.

S o l . v = 2 () [L2 – L1] = 2 × 320 [73 – 20] × 10–2 ms–1

= 339 m/s 27.



t  m as ‘e’ is same for electron and proton.

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s2 at a distance of 5 m from the mean position. The time period of oscillation is (1) 2 s

(2) 2 s

(3)  s

(4) 1 s

S o l . |a| = 2y  20 = 2(5)   = 2 rad/s

(1) Independent of the distance between the plates

2 2  s  2 The power radiated by a black body is P and it radiates maximum energy at wavelength, 0. If the temperature of the black body is now changed so that it radiates maximum energy 3 at wavelength  0 , the power radiated by it 4 becomes nP. The value of n is T

(2) Proportional to the square root of the distance between the plates

30.

(3) Linearly proportional to the distance between the plates (4) Inversely proportional to the distance between the plates Answer ( 1 ) S o l . For isolated capacitor Q = Constant

Q2 2A0

F is Independent of the distance between plates. 28.

t

Answer ( 3 )

The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is

Fplate 

2hm eE



∵ Electron has smaller mass so it will take smaller time.

Answer ( 3 )

= 339.2

1 eE 2 t 2 m

An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is

(1)

3 4

(2)

256 81

(3)

4 3

(4)

81 256

Answer ( 2 ) S o l . We know,

max T  constant (Wien's law)

So, max1 T1  max2 T2  0 T 

(1) Smaller (2) 10 times greater

 T 

(3) 5 times greater

3 0 T 4

4 T 3 4

(4) Equal

So,

Answer ( 1 ) 8

4

P2  T   256 4      P1  T  81 3


31.

Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by l on applying a force F, how much force is needed to stretch the socond wire by the same amount?

33.

A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 105 Nm–2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is (1) 104.3 J

(1) 9 F

(2) 42.2 J

(2) 4 F

(3) 208.7 J

(3) 6 F

(4) 84.5 J

(4) F Answer ( 1 )

Answer ( 3 )

S o l . Wire 1 :

S o l . Q = U + W A, 3l

 54 × 4.18 = U + 1.013 × 105(167.1 × 10–6 – 0)

F

 U = 208.7 J

Wire 2 : 3A, l

34.

F

For wire 1,

 F  l    3l  AY 

(1) 8iˆ  4 ˆj  7kˆ …(i) (2) 7iˆ  8ˆj  4kˆ

For wire 2,

(3) 4iˆ  ˆj  8kˆ

F l Y 3A l

 F   l   l  3AY 

The moment of the force, F  4iˆ  5 ˆj  6kˆ at (2, 0, –3), about the point (2, –2, –2), is given by

(4) 7iˆ  4 ˆj  8kˆ …(ii) Answer ( 4 )

From equation (i) & (ii),

Sol.

Y

 F   F  l   3l    l  AY   3AY   32.

F A

F  9F

A small sphere of radius 'r' falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to (1) r3

(2) r5

(3) r2

(4) r4

r  r0

r0

P

r

O   (r  r0 )  F

X ...(i)

ˆ  (2iˆ  2 ˆj  2k) ˆ r  r0  (2iˆ  0ˆj  3k)

Answer ( 2 )

 0iˆ  2 ˆj  kˆ

2 S o l . Power = 6 rVT iVT  6 rVT

î ˆj kˆ   0 2 1  7iˆ  4 ˆj  8kˆ 4 5 6

VT  r 2  Power  r 5 9


35.

A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of –0.004 cm, the correct diameter of the ball is

...(i)

N cos  = mg

...(ii)

tan  

a g

a = g tan 

(1) 0.521 cm

37.

= 0.5 cm + 25 × 0.001 – (–0.004)

A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E . Due to the force q E , its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively

= 0.5 + 0.025 + 0.004

(1) 2 m/s, 4 m/s

= 0.529 cm

(2) 1 m/s, 3.5 m/s

(2) 0.053 cm (3) 0.525 cm (4) 0.529 cm Answer ( 4 ) S o l . Diameter of the ball = MSR + CSR × (Least count) – Zero error

36.

N sin  = ma

A block of mass m is placed on a smooth inclined wedge ABC of inclination  as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and  for the block to remain stationary on the wedge is

(3) 1 m/s, 3 m/s (4) 1.5 m/s, 3 m/s Answer ( 3 ) Sol. t = 0

A

A

m

a

v = 6 ms C t=3

v=0

a B

(1) a 

g cosec 

(2) a = g cos 

(3) a 

g sin 

(4) a = g tan 

Acceleration a 

S1 

–a –1

60  6 ms2 1

1  6(1)2 = 3 m 2

...(i)

For t = 1 s to t = 2 s,

N cos

S2  6.1 

N

1  6(1)2  3 m 2

...(ii)

For t = 2 s to t = 3 s,

 ma (pseudo)

t=2 B v=0

For t = 0 to t = 1 s,

Answer ( 4 ) Sol.

–1

v = –6 ms

 C

–a

t=1

N sin

S3  0 



1  6(1)2  3 m 2

...(iii)

Total displacement S = S1 + S2 + S3 = 3 m

mg



a Average velocity 

In non-inertial frame, 10

3  1 ms 1 3


Total distance travelled = 9 m Average speed 

38.

Sol.

9  3 ms 1 3

h

B

A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

As track is frictionless, so total mechanical energy will remain constant

(1) 0.5

0  mgh 

A

vL

T.M.EI =T.M.EF

(2) 0.8 h

(3) 0.25

1 mvL2  0 2

vL2 2g

(4) 0.4 For completing the vertical circle, vL  5gR

Answer ( 3 ) S o l . According to law of conservation of linear momentum,

h

mv  4m  0  4mv  0

v 

40.

v 4

v Relative velocity of separation 4 e  Relative velocity of approach v e 39.

5gR 5 5  R D 2g 2 4

Three objects, A : (a solid sphere), B : (a thin circular disk) and C : (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed  about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation (1) WC > WB > WA

1  0.25 4

(2) WB > WA > WC

A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to

(3) WA > WB > WC (4) WA > WC > WB Answer ( 1 ) S o l . Work done required to bring them rest

h

W = KE

B

W 

A

(1)

W  I for same 

3 D 2

WA : WB : WC 

7 (2) D 5

(3) D (4)

1 2 I 2

=

5 D 4

2 1 MR2 : MR2 : MR2 5 2 2 1 : :1 5 2

= 4 : 5 : 10  WC > WB > WA

Answer ( 4 ) 11


41.

S o l . ex = 0

Which one of the following statements is incorrect? (1) Rolling friction is smaller than sliding friction.

dL 0 dt i.e. L = constant

(2) Frictional force opposes the relative motion.

So angular momentum remains constant.

So,

44.

(3) Limiting value of static friction is directly proportional to normal reaction. (4) Coefficient of sliding dimensions of length.

friction

has

(1) Raindrops will fall faster

Answer ( 4 )

(2) Time period of a simple pendulum on the Earth would decrease

S o l . Coefficient of sliding friction has no dimension.

(3) Walking on the ground would become more difficult

f = sN  s 

42.

If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?

f N

(4) ‘g’ on the Earth will not change Answer ( 4 )

A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy (K t ) as well as rotational kinetic energy (K r ) simultaneously. The ratio Kt : (Kt + Kr) for the sphere is

S o l . If Universal Gravitational constant becomes ten times, then G = 10 G

(1) 7 : 10

(2) 10 : 7

45.

(3) 5 : 7

(4) 2 : 5

So, acceleration due to gravity increases. i.e. (4) is wrong option.

Answer ( 3 ) S o l . Kt 

1 mv 2 2

Kt  Kr 

1 1 1 1 2  v  mv2  I2  mv2   mr 2   2 2 2 25  r  

The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB and KC, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then

2

B A

C

S

7 mv2 10

(1) KA < KB < KC So, 43.

Kt 5  Kt  Kr 7

(2) KB < KA < KC (3) KA > KB > KC

A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?

(4) KB > KA > KC Answer ( 3 ) B

Sol. perihelion A

(1) Angular velocity

S VA

(2) Rotational kinetic energy

VC C aphelion

(3) Moment of inertia

Point A is perihelion and C is aphelion.

(4) Angular momentum

So, VA > VB > VC So, KA > KB > KC

Answer ( 4 ) 12


46.

48.

Given van der Waals constant for NH3, H2, O2 and CO2 are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquefied?

(Given molar mass of BaSO4 = 233 g mol–1)

(1) NH3

(1) 1.08 × 10–10 mol2L–2

(2) O2

(2) 1.08 × 10–14 mol2L–2

(3) H2

(3) 1.08 × 10–12 mol2L–2

(4) CO2

(4) 1.08 × 10–8 mol2L–2

Answer ( 1 ) Sol. • • 47.

Answer ( 1 )

van der waal constant ‘a’, signifies intermolecular forces of attraction.

S o l . Solubility of BaSO4, s =

Higher is the value of ‘a’, easier will be the liquefaction of gas.

Ba2  (aq)  SO 24(aq) BaSO 4 (s) s

= (1.04 × 10–5)2 = 1.08 × 10–10 mol2 L–2

M M HCl + 45 mL NaOH b. 55 mL 10 10

49.

M M HCl + 25 mL NaOH 5 5

On which of the following properties does the coagulating power of an ion depend? (1) The magnitude of the charge on the ion alone (2) Both magnitude and sign of the charge on the ion

M M HCl + 100 mL NaOH 10 10

(3) Size of the ion alone

pH of which one of them will be equal to 1?

(4) The sign of charge on the ion alone

(1) b (2) d

Answer ( 2 )

(3) a

Sol. •

(4) c Answer ( 4 ) Sol. •

•

•

1 Meq of NaOH = 25   1 = 5 5

•

Meq of HCl in resulting solution = 10

•

Molarity of [H+] in resulting mixture

Coagulation of colloidal solution by using an electrolyte depends on the charge present (positive or negative) on colloidal particles as well as on its size.

1 Meq of HCl = 75   1 = 15 5

=

s

Ksp = [Ba2+] [SO42–]= s2

M M HCl + 40 mL NaOH a. 60 mL 10 10

d. 100 mL

2.42  103 (mol L–1) 233

= 1.04 × 10–5 (mol L–1)

Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations :

c. 75 mL

The solubility of BaSO4 in water is 2.42 × 10–3 gL–1 at 298 K. The value of its solubility product (Ksp) will be

Coagulating power of an electrolyte depends on the magnitude of charge present on effective ion of electrolyte.

50. Considering Ellingham diagram, which of the following metals can be used to reduce alumina? (1) Fe

10 1  100 10

(2) Mg

 1 pH = –log[H+] =  log   = 1.0  10 

(3) Zn (4) Cu 13


54. Which of the following statements is not true for halogens?

Answer ( 2 ) S o l . The metal which is more reactive than 'Al' can reduce alumina i.e. 'Mg' should be the correct option.

(1) All form monobasic oxyacids (2) All but fluorine show positive oxidation states

51. The correct order of atomic radii in group 13 elements is

(3) All are oxidizing agents

(1) B < Al < In < Ga < Tl

(4) Chlorine has the highest electron-gain enthalpy

(2) B < Ga < Al < Tl < In (3) B < Al < Ga < In < Tl

Answer ( 2 )

(4) B < Ga < Al < In < Tl

S o l . Due to high electronegativity and small size, F forms only one oxoacid, HOF known as Fluoric (I) acid. Oxidation number of F is +1 in HOF.

Answer ( 4 ) Sol. Elements Atomic radii (pm)

B 85

Ga 135

Al 143

In 167

55. Which one of the following elements is unable to form MF63– ion?

Tl 170

(1) Ga

52. In the structure of ClF3, the number of lone pairs of electrons on central atom ‘Cl’ is

(2) B

(1) One

(3) Al

(2) Four

(4) In

(3) Two

Answer ( 2 )

(4) Three

S o l . ∵ 'B' has no vacant d-orbitals in its valence shell, so it can't extend its covalency beyond 4. i.e. 'B' cannot form the ion like MF6 3(–) i.e. BF63(–).

Answer ( 3 )

 

F

 

 

Hence, the correct option is (2). 56.

 

F

 

Cl

F

 

 

 

S o l . The structure of ClF3 is

Regarding cross-linked or network polymers, which of the following statements is incorrect? (1) They contain covalent bonds between various linear polymer chains.

 

The number of lone pair of electrons on central Cl is 2.

(2) Examples are bakelite and melamine. (3) They are formed from bi- and tri-functional monomers.

53. The correct order of N-compounds in its decreasing order of oxidation states is

(4) They contain strong covalent bonds in their polymer chains.

(1) HNO3, NO, N2, NH4Cl (2) HNO3, NH4Cl, NO, N2

Answer ( 4 )

(3) HNO3, NO, NH4Cl, N2

S o l . Cross linked or network polymers are formed from bi-functional and tri-functional monomers and contain strong covalent bonds between various linear polymer chains, e.g. bakelite, melamine etc. Option (4) is not related to cross-linking.

(4) NH4Cl, N2, NO, HNO3 Answer ( 1 ) 5

2

0

–3

S o l . H N O , N O, N2 , NH Cl 3 4

So option (4) should be the correct option.

Hence, the correct option is (1). 14


57.

S o l . Amylose and Amylopectin are polymers of D-glucose, so -link is not possible. Amylose is linear with 1  4 -linkage whereas Amylopectin is branched and has both 1  4 and 1  6 -linkages.

Nitration of aniline in strong acidic medium also gives m-nitroaniline because (1) Inspite of substituents nitro group always goes to only m-position. (2) In absence of substituents nitro group always goes to m-position.

So option (1) should be the correct option. 60.

(3) In electrophilic substitution reactions amino group is meta directive. (4) In acidic (strong) medium aniline is present as anilinium ion. Answer ( 4 )

NH2

NH3

(3) 3.0

(4) 4.4

–NH3 is m-directing, hence besides para

COOH

(51%) and ortho (2%), meta product (47%) is also formed in significant yield.

COOH

Which of the following oxides is most acidic in nature? (1) MgO

(2) BaO

(3) BeO

(4) CaO

Conc.H2SO4

CO(g) + CO2 (g) + H2O(l) 1 mol 20

1 mol 20

 1  4.5 g or  mol   20 

Gaseous mixture formed is CO and CO2 when it is passed through KOH, only CO 2 is absorbed. So the remaining gas is CO. So, weight of remaining gaseous product CO is

Answer ( 3 ) 

2  28  2.8 g 20

Basic character increases.

So, the correct option is (2)

S o l . BeO < MgO < CaO < BaO

So, the most acidic should be BeO. In fact, BeO is amphoteric oxide while other given oxides are basic. 59.

(2) 2.8

Conc.H2 SO4 S o l . HCOOH   CO(g)  H2 O(l) 1 1   mol 2.3 g or  mol  20  20 

Anilinium ion

58.

(1) 1.4

Answer ( 2 )

H

Sol.

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H2SO4. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

The difference amylopectin is

between

amylose

61.

In the reaction

O–Na+

OH

and

CHO

+ CHCl3 + NaOH

(1) Amylopectin have 1  4 -linkage and 1 6 -linkage

The electrophile involved is

(2) Amylopectin have 1  4 -linkage and 1  6 -linkage

(1) Dichloromethyl cation CHCl2

(3) Amylose have 1  4 -linkage and 1  6 -linkage

(2) Dichloromethyl anion CHCl2

(4) Amylose is made up of glucose and galactose

(3) Formyl cation CHO















(4) Dichlorocarbene : CCl2 

Answer ( 1 ) 15






S o l . Option (2) is secondary alcohol which on oxidation gives phenylmethyl ketone (Acetophenone). This on reaction with I2 and NaOH form iodoform and sodium benzoate.

Answer ( 4 ) S o l . It is Reimer-Tiemann reaction. The electrophile formed is :CCl2 (Dichlorocarbene) according to the following reaction

2NaOH  I2  NaOI  NaI  H2 O

.–. CCl3  H2 O CHCl3  OH –

.–.

CCl3   : CCl2  Cl

CH – CH3

NaOI

OH (A)

–

C – CH3 O Acetophenone

Electrophile

I2

COONa + CHI3

62.

Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their

Sodium benzoate

64.

(1) Formation of intramolecular H-bonding

Which of the following molecules represents the order of hybridisation sp2, sp2, sp, sp from left to right atoms?

(2) More extensive association of carboxylic acid via van der Waals force of attraction

(1) HC  C – C  CH

(3) Formation of carboxylate ion

(3) CH2 = CH – C  CH

(4) Formation of intermolecular H-bonding

(4) CH3 – CH = CH – CH3

(2) CH2 = CH – CH = CH2

Answer ( 3 )

Answer ( 4 )

sp2

S o l . Due to formation of intermolecular H-bonding in carboxylic acid, association occurs. Hence boiling point increases and become more than the boiling point of aldehydes, ketones and alcohols of comparable molecular masses. 63.

Iodoform (Yellow PPt)

NaOH

sp2

sp

sp

S o l . CH2  CH – C  CH Number of orbital require in hybridization = Number of -bonds around each carbon atom. 65.

Compound A, C8H10O, is found to react with NaOI (produced by reacting Y with NaOH) and yields a yellow precipitate with characteristic smell.

Which of the following carbocations is expected to be most stable?

NO2

NO2

A and Y are respectively (1) (1) H3C

(2)

CH2 – OH and I2

 Y

H



CH2 – CH2 – OH and I2

H



(4) Y

Y

CH3



NO2

(3)

(4) CH3

Y

NO2

CH – CH3 and I2 OH

(3)

(2) H

H

Answer ( 2 ) S o l . –NO2 group exhibit –I effect and it decreases with increase in distance. In option (2) positive charge present on C-atom at maximum distance so –I effect reaching to it is minimum and stability is maximum.

OH and I2

Answer ( 2 ) 16


66.

Which of the following is correct with respect to – I effect of the substituents? (R = alkyl)

69.

(1) – NH2 < – OR < – F (2) – NH2 > – OR > – F

Iron carbonyl, Fe(CO)5 is (1) Tetranuclear

(2) Trinuclear

(3) Mononuclear

(4) Dinuclear

Answer ( 3 )

(3) – NR2 < – OR < – F

S o l . • Based on the number of metal atoms present in a complex, they are classified into mononuclear, dinuclear, trinuclear and so on.

(4) – NR2 > – OR > – F Answer ( 1 * ) S o l . –I effect increases on increasing electronegativity of atom. So, correct order of –I effect is –NH2 < – OR < – F.

eg: Fe(CO)5 : mononuclear Co2(CO)8 : dinuclear

*Most appropriate Answer is option (1), however option (3) may also be correct answer. 67.

Fe3(CO)12: trinuclear Hence, option (3) should be the right answer.

The type of isomerism shown by the complex [CoCl2(en)2] is 70.

(1) Geometrical isomerism (2) Ionization isomerism

Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code :

(3) Coordination isomerism

Column I

(4) Linkage isomerism

Column II

a. Co3+

i.

8 BM

b. Cr3+

ii.

35 BM

c. Fe3+

iii.

3 BM

d. Ni2+

iv.

24 BM

v.

15 BM

Answer ( 1 ) S o l . In [CoCl2(en)2], Coordination number of Co is 6 and this compound has octahedral geometry.

• As per given option, type of isomerism is geometrical isomerism. 68.

Which one of the following ions exhibits d-d transition and paramagnetism as well? (1)

CrO42–

(2)

(3) Cr2O72–

v

ii

i

(2)

iv

i

ii

iii

(3)

i

ii

iii

iv

(4)

iii

v

i

ii

4(4  2)  24 BM

3(3  2)  15 BM

= [Ar] Fe3+ = [Ar] 3d5, Unpaired e–(n) = 5

MnO42– = Mn6+ = [Ar] 3d1

Spin magnetic moment =

Unpaired electron (n) = 1; Paramagnetic =

iv


Unpaired electron (n) = 0; Diamagnetic

Mn7+

(1)

Cr3+ = [Ar] 3d3, Unpaired e–(n) = 3

Unpaired electron (n) = 0; Diamagnetic

MnO4–

d


S o l . CrO42–  Cr6+ = [Ar] 

c

S o l . Co3+ = [Ar] 3d6, Unpaired e–(n) = 4

(4) MnO42–

Cr6+

b

Answer ( 1 )

MnO4–

Answer ( 4 )

Cr2O72–

a

5(5  2)  35 BM

Ni2+ = [Ar] 3d8, Unpaired e–(n) = 2

= [Ar] Spin magnetic moment =

Unpaired electron (n) = 0; Diamagnetic 17

2(2  2)  8 BM


71.

The geometry and magnetic behaviour of the complex [Ni(CO)4] are

Sol. 

(1) Square planar geometry and diamagnetic (2) Square planar paramagnetic

geometry

For first order reaction, t 1/2  which is independent concentration of reactant.

and 

(3) Tetrahedral geometry and diamagnetic (4) Tetrahedral geometry and paramagnetic

of

For second order reaction, t 1/2 

0.693 , k

initial

1 , k[A0 ]

which depends on initial concentration of reactant.

Answer ( 3 ) 73.

S o l . Ni(28) : [Ar]3d8 4s2 ∵ CO is a strong field ligand

Among CaH2, BeH2, BaH2, the order of ionic character is (1) BeH2 < CaH2 < BaH2

Configuration would be :

(2) BeH2 < BaH2 < CaH2

3

sp -hybridisation ××

×× ×× ××

CO

CO CO CO

(3) CaH2 < BeH2 < BaH2 (4) BaH2 < BeH2 < CaH2 Answer ( 1 )

For, four ‘CO’-ligands hybridisation would be sp 3 and thus the complex would be diamagnetic and of tetrahedral geometry.

S o l . For 2nd group hydrides, on moving down the group metallic character of metals increases so ionic character of metal hydride increases.

CO

Hence the option (1) should be correct option. 74.

Ni

CO

OC

(1) 18 mL of water (2) 0.00224 L of water vapours at 1 atm and 273 K

CO 72.

In which case is number of molecules of water maximum?

The correct difference between first and second order reactions is that

(3) 0.18 g of water (4) 10–3 mol of water

(1) The rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations

Answer ( 1 ) S o l . (1) Mass of water = 18 × 1 = 18 g

(2) A first-order reaction can catalyzed; a second-order reaction cannot be catalyzed

Molecules of water = mole × NA =

18 NA 18

= NA

(3) The half-life of a first-order reaction does not depend on [A] 0 ; the half-life of a second-order reaction does depend on [A]0

(2) Moles of water =

(4) The rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations

(3) Molecules of water = mole × NA =

0.00224 = 10–4 22.4

Molecules of water = mole × NA = 10–4 NA 0.18 NA 18

= 10–2 NA

Answer ( 3 )

(4) Molecules of water = mole × NA = 10–3 NA 18


75.

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :

– BrO4

– BrO3

1.82 V

– Br

1.0652 V

1.5 V

Br2

On solving, we get



HBrO

 X = 800 kJ/mole 77.

1.595 V

Then the species disproportionation is

undergoing

When initial concentration of the reactant is doubled, the half-life period of a zero order reaction (1) Is halved

(1) BrO3

(2) Is tripled

(2) Br2

(3) Is doubled

(3) BrO4

(4) Remains unchanged Answer ( 3 )

(4) HBrO

S o l . Half life of zero order

Answer ( 4 ) 1

0

o S o l . HBrO   Br2 , EHBrO/Br  1.595 V

t 1/2 

2

1

5

HBrO   BrO3 , Eo

BrO3 /HBrO

[A0 ] 2K

t 1/2 will be doubled on doubling the initial concentration.

 1.5 V

o for the disproportionation of HBrO, Ecell

78.

For the redox reaction

o o Ecell  EHBrO/Br  Eo

MnO4  C2 O24  H   Mn2   CO2  H2 O

= 1.595 – 1.5 = 0.095 V = + ve Hence, option (4) is correct answer.

The correct coefficients of the reactants for the balanced equation are

2

76.

X X   200 2 4

BrO3 /HBrO

C2 O24

H+

(1) 16

5

2

(2) 2

16

5

(1) 200 kJ mol–1

(3) 2

5

16

(2) 800 kJ mol–1

(4) 5

16

2

The bond dissociation energies of X2, Y2 and XY are in the ratio of 1 : 0.5 : 1. H for the formation of XY is –200 kJ mol–1. The bond dissociation energy of X2 will be

MnO4

(3) 100 kJ mol–1

Answer ( 3 )

Reduction

(4) 400 kJ mol–1 +7

S o l . MnO4– + C2O42– + H+

Answer ( 2 ) S o l . The reaction for fH°(XY)

+4

 n-factor of MnO4  5

n-factor of C2 O24  2

X , X 2

 Ratio of n-factors of MnO4 and C2 O24 is 5 : 2 So, molar ratio in balanced reaction is 2 : 5

respectively 

2+

Mn + CO2 + H2O

Oxidation

1 1  XY(g) X2 (g)  Y2 (g)  2 2

Bond energies of X2, Y2 and XY are X,

+3

 The balanced equation is

X X H      X  200 2 4

2MnO4  5C2 O24  16H  2Mn2   10CO2  8H2 O 19


79.

Answer ( 4 )

Which one of the following conditions will favour maximum formation of the product in the reaction,

S o l . Element (X) electronic configuration 1s2 2s2 2p3

X2 (g) r H   X kJ? A2 (g)  B2 (g)

So, valency of X will be 3.

(1) Low temperature and high pressure

Valency of Mg is 2.

(2) High temperature and high pressure

Formula of compound formed by Mg and X will be Mg3X2.

(3) Low temperature and low pressure 82. (4) High temperature and low pressure Answer ( 1 ) X2 (g); H  x kJ S o l . A2 (g)  B2 (g) On increasing pressure equilibrium shifts in a direction where pressure decreases i.e. forward direction.

(1)

3 2

On decreasing temperature, equilibrium shifts in exothermic direction i.e., forward direction.

(2)

3 3 4 2

So, high pressure and low temperature favours maximum formation of product. 80.

Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is

(3)

The correction factor ‘a’ to the ideal gas equation corresponds to (4)

(1) Density of the gas molecules

4 3 3 2

1 2

(2) Electric field present between the gas molecules

Answer ( 2 )

(3) Volume of the gas molecules

S o l . For BCC lattice : Z = 2, a 

(4) Forces of attraction between the gas molecules

3

For FCC lattice : Z = 4, a = 2 2 r

Answer ( 4 ) 2   S o l . In real gas equation,  P  an  (V  nb)  nRT  V2   van der Waal’s constant, ‘a’ signifies intermolecular forces of attraction.

81.

4r



d25C d900C





Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is 1s2 2s2 2p3, the simplest formula for this compound is

83.

 ZM   N a3   A   ZM   N a3   A 

BCC

FCC 3

3 3 2  2 2 r    4r  4 2  4    3 

Consider the following species : CN+, CN–, NO and CN

(1) Mg2X3 (2) Mg2X

Which one of these will have the highest bond order?

(3) MgX2

(1) NO

(2) CN+

(4) Mg3X2

(3) CN–

(4) CN

20


Answer ( 3 )

85.

S o l . NO : (1s) 2 , ( 1s) 2 , (2s) 2,( 2s) 2,(2p z) 2 , (2px)2 = (2py)2,(2px)1 = (2py)0 BO =

10  5  2.5 2

CN–

(1s)2,

:

(1s)2,

(1) N2O5 (2) N2O (2s)2,(2s)2,

(2px)2

(3) NO2

= (2py)2,(2pz)2

(4) NO

10  4 3 2

Answer ( 1 )

BO =

S o l . Fact

CN : (1s)2, (1s)2, (2s)2,(2s)2, (2px)2

86.

= (2py)2,(2pz)1 BO =

9 4  2.5 2

CN+ : (1s)2, (1s)2, (2s)2,(2s)2, (2px)2

Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is (1) CH  CH

= (2py)2

84.

Which oxide of nitrogen is not a common pollutant introduced into the atmosphere both due to natural and human activity?

(2) CH3 – CH3

8 4 2 BO = 2

(3) CH2  CH2

Hence, option(3) should be the right answer.

(4) CH4

Which one is a wrong statement?

Answer ( 4 )

(1) Total orbital angular momentum of electron in 's' orbital is equal to zero

S o l . CH4 (A)

Br2/h

CH3Br Na/dry ether Wurtz reaction

(2) The electronic configuration of N atom is 1s2

1

2s2

2px

1

2py

1

2pz

CH3 — CH3 Hence the correct option is (4)

(3) An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers

87.

The compound A on treatment with Na gives B, and with PCl 5 gives C. B and C react together to give diethyl ether. A, B and C are in the order

(4) The value of m for dz2 is zero

(1) C2H5OH, C2H6, C2H5Cl

Answer ( 2 )

(2) C2H5Cl, C2H6, C2H5OH

S o l . According to Hund's Rule of maximum multiplicity, the correct electronic configuration of N-atom is

(3) C2H5OH, C2H5Cl, C2H5ONa (4) C2H5OH, C2H5ONa, C2H5Cl Answer ( 4 )

1s2

2s2

S o l . C2H5OH (A)

2p3

Na

C2H5O Na+ (B)

PCl5

OR

C2H5Cl (C)

2

1s

2s

2

2p

C2H5O Na+ + C2H5Cl (B) (C)

3

 Option (2) violates Hund's Rule.

SN2

C2H5OC2 H5

So the correct option is (4) 21


88.

CH2CH2CH3

The compound C7H8 undergoes the following reactions:

CHO

(3) 3Cl / 

Br /Fe

,

,

Zn/HCl

2 2 C7H8   A   B  C

OH

CH(CH3)2 ,

(4)

The product 'C' is (1) m-bromotoluene

,

Cl S o l . CH CH CH – Cl + 3 2 2

(3) o-bromotoluene

Al Cl

Cl

(4) p-bromotoluene

1, 2–H Shift

+

CH3 – CH – CH3

Answer ( 1 )

+

CH3CH2CH2

Cl

(Incipient carbocation)

–

AlCl3

Cl

CH3

CCl3 3Cl 2 

(C7H8)

CH3 – CO – CH3

Answer ( 4 )

(2) 3-bromo-2,4,6-trichlorotoluene

Sol.

COOH

CCl3 Br2 Fe

(B)

(A)

–

AlCl3

Now,

Br

CH3

Zn HCl

CH – CH3 O2

CH3 – CH – CH3

CH3

(P) (C)

Br

CH3

So, the correct option is (1)

+

H /H2O

CH3 – C – CH3 + 89.

Identify the major products P, Q and R in the following sequence of reactions:

(R) 90.

+ CH3CH2CH2Cl P P

CH(CH3)2

(2)

(i) O2

,

Which of the following compounds can form a zwitterion?

(2) Benzoic acid

Q+R

(ii) H3O+/

(3) Acetanilide R

(4) Glycine

CHO ,

(Q)

Hydroperoxide Rearrangement

(1) Aniline

Q

CH2CH2CH3

(1)

Anhydrous AlCl3

HC –C – O– O –H 3

OH

O

Answer ( 4 ) CH3CH2 – OH

,

Sol.

OH



H3N – CH2 – COOH pKa = 9.60

, CH3CH(OH)CH3



H3N – CH2 – COO– (Zwitterion form)

pKa = 2.34

H2N – CH2 – COO–

22


91.

Which of the following statements is correct?

95.

(1) Ovules are not enclosed by ovary wall in gymnosperms

Casparian strips occur in (1) Epidermis (2) Cortex

(2) Horsetails are gymnosperms

(3) Pericycle

(3) Selaginella is heterosporous, while Salvinia is homosporous

(4) Endodermi Answer ( 4 )

(4) Stems are usually unbranched in both Cycas and Cedrus

Sol. •

Answer ( 1 ) Sol. • • 92.

Gymnosperms have naked ovule.

•

Called phanerogams without womb/ovary

96.

Secondary xylem and phloem in dicot stem are produced by

(1) Halophytes

(1) Apical meristems

(2) Carnivorous plants

(2) Phellogen

(3) Free-floating hydrophytes

(3) Vascular cambium

(4) Submerged hydrophytes

(4) Axillary meristems

Sol.  

Halophytes like pneumatophores.

Answer ( 3 ) mangrooves

have

Sol. •

Apogeotropic (–vely geotropic) roots having lenticels called pneumathodes to uptake O2.

Sweet potato is a modified (1) Stem

97.

(2) Tap root

Vascular cambium is partially secondary

•

Form secondary xylem towards its inside and secondary phloem towards outsides.

•

4 – 10 times more secondary xylem is produced than secondary phloem.

Select the wrong statement : (1) Cell wall is present in members of Fungi and Plantae

(3) Adventitious root (4) Rhizome

(2) Pseudopodia are locomotory and feeding structures in Sporozoans

Answer ( 3 ) S o l . Sweet potato is a modified adventitious root for storage of food

94.

It is suberin rich.

Pneumatophores occur in

Answer (1)

93.

Endodermis have casparian strip on radial and inner tangential wall.

•

Rhizomes are underground modified stem

•

Tap root is primary root directly elongated from the redicle

(3) Mushrooms belong to Basidiomycetes (4) Mitochondria are the powerhouse of the cell in all kingdoms except Monera Answer ( 2 )

Plants having little or no secondary growth are (1) Grasses

S o l . Pseudopodia are locomotory structures in sarcodines (Amoeboid)

(2) Conifers

98.

(3) Deciduous angiosperms

The experimental proof for semiconservative replication of DNA was first shown in a

(4) Cycads

(1) Fungus

(2) Plant

(3) Bacterium

(4) Virus

Answer (1)

Answer ( 3 )

S o l . Grasses are monocots and monocots usually do not have secondary growth. Palm like monocots secondary growth.

have

S o l . Semi-conservative DNA replication was first shown in Bacterium Escherichia coli by Matthew Meselson and Franklin Stahl.

anomalous 23

NEET (UG) - 2018 (Code-EE) CHLAAA

99.

S o l . Starch synthesis in pea is controlled by pleiotropic gene.

Select the correct match (1) Alec Jeffreys

- Streptococcus

Other options (2, 3 & 4) are correctly matched.

pneumoniae (2) Matthew Meselson

- Pisum sativum

102. Offsets are produced by

and F. Stahl (3) Alfred Hershey and

(1) Meiotic divisions - TMV

(2) Parthenocarpy

Martha Chase

(3) Mitotic divisions

(4) Francois Jacob and - Lac operon

(4) Parthenogenesis

Jacques Monod

Answer ( 3 )

Answer ( 4 )

S o l . Offset is a vegetative part of a plant, formed by mitosis.

S o l . Francois Jacob and Jacque Monod proposed model of gene regulation known as operon model/lac operon. –

Alec Jeffreys – DNA fingerprinting technique.

–

Matthew Meselson and F. Stahl – Semiconservative DNA replication in E. coli.

–

Alfred Hershey and Martha Chase – Proved DNA as genetic material not protein

–

Meiotic divisions do not occur in somatic cells.

–

Parthenogenesis is the formation of embryo from ovum or egg without fertilisation.

–

Parthenocarpy is the fruit formed without fertilisation, (generally seedless)

103. Which of the following flowers only once in its life-time? (1) Bamboo species

100. Select the correct statement (1) Franklin Stahl coined the term ‘‘linkage’’

(2) Mango

(2) Spliceosomes take part in translation

(3) Jackfruit

(3) Punnett square was developed by a British scientist

(4) Papaya Answer ( 1 )

(4) Transduction was discovered by S. Altman

S o l . Bamboo species are monocarpic i.e., flower generally only once in its life-time after 50100 years.

Answer ( 3 ) S o l . Punnett square was developed by a British geneticist, Reginald C. Punnett. –

Franklin Stahl proved semi-conservative mode of replication.

–

Transduction was discovered by Zinder and Laderberg.

–

Spliceosome formation is part of posttranscriptional change in Eukaryotes

Jackfruit, papaya and mango are polycarpic i.e., produce flowers and fruits many times in their life-time. 104. Which of the following has proved helpful in preserving pollen as fossils?

: Grasshopper

(4) T.H. Morgan

: Linkage

(4) Sporopollenin

Pollenkitt – Help in insect pollination.

determination : Co-dominance

(3) Cellulosic intine

S o l . Sporopollenin cannot be degraded by enzyme; strong acids and alkali, therefore it is helpful in preserving pollen as fossil.

(1) Starch synthesis in pea : Multiple alleles

(3) ABO blood grouping

(2) Oil content

Answer ( 4 )

101. Which of the following pairs is wrongly matched? (2) XO type sex

(1) Pollenkitt

Cellulosic Intine – Inner sporoderm layer of pollen grain known as intine made up cellulose & pectin. Oil content – No role is pollen preservation.

Answer ( 1 ) 24


105. Which of the following is commonly used as a vector for introducing a DNA fragment in human lymphocytes?

108. Use of bioresources by multinational companies and organisations without authorisation from the concerned country and its people is called

(1) Retrovirus (2)  phage

(1) Bio-infringement

(3) Ti plasmid

(2) Biodegradation

(4) pBR 322

(3) Biopiracy (4) Bioexploitation

Answer ( 1 )

Answer ( 3 )

S o l . Retrovirus is commonly used as vector for introducing a DNA fragment in human lymphocyte.

S o l . Biopiracy is term used for or refer to the use of bioresources by multinational companies and other organisation without proper authorisation from the countries and people concerned with compensatory payment (definition of biopiracy given in NCERT).

Gene therapy : Lymphocyte from blood of patient are grown in culture outside the body, a functional gene is introduced by using a retroviral vector, into these lymphocyte. 106. The correct order of steps in Polymerase Chain Reaction (PCR) is

109. A ‘new’ variety of rice was patented by a foreign company, though such varieties have been present in India for a long time. This is related to

(1) Extension, Denaturation, Annealing (2) Denaturation, Extension, Annealing (3) Annealing, Extension, Denaturation

(1) Co-667

(4) Denaturation, Annealing, Extension

(2) Lerma Rojo

Answer ( 4 )

(3) Sharbati Sonora

S o l . This technique is used for making multiple copies of gene (or DNA) of interest in vitro.

(4) Basmati Answer ( 4 )

Each cycle has three steps

S o l . In 1997, an American company got patent rights on Basmati rice through the US patent and trademark office that was actually been derived from Indian farmer’s varieties.

(I) Denaturation (II) Primer annealing (III) Extension of primer

The diversity of rice in India is one of the richest in the world, 27 documented varieties of Basmati are grown in India.

107. In India, the organisation responsible for assessing the safety of introducing genetically modified organisms for public use is

Indian basmati was crossed with semi-dwarf varieties and claimed as an invention or a novelty.

(1) Indian Council of Medical Research (ICMR) (2) Research Committee Manipulation (RCGM)

on

Genetic

Sharbati Sonora and Lerma Rojo are varieties of wheat.

(3) Council for Scientific and Industrial Research (CSIR)

110. Select the correct match

(4) Genetic Engineering Appraisal Committee (GEAC) Answer ( 4 )

(1) Ribozyme

- Nucleic acid

(2) T.H. Morgan

- Transduction

(3) F2 × Recessive parent - Dihybrid cross

S o l . Indian Government has setup organisation such as GEAC (Genetic Engineering Appraisal Committee) which will make decisions regarding the validity of GM research and safety of introducing GM-organism for public services. (Direct from NCERT).

(4) G. Mendel

- Transformation

Answer ( 1 ) S o l . Ribozyme is a catalytic RNA, which is nucleic acid. 25


115. In stratosphere, which of the following elements acts as a catalyst in degradation of ozone and release of molecular oxygen?

111. Niche is (1) all the biological factors in the organism's environment (2) the range of temperature that the organism needs to live (3) the physical space where an organism lives

(1) Carbon

(2) Fe

(3) Cl

(4) Oxygen

Answer ( 3 ) S o l . UV rays act on CFCs, releasing Cl atoms, chlorine reacts with ozone in sequential method converting into oxygen

(4) the functional role played by the organism where it lives Answer ( 4 )

Carbon, oxygen and Fe are not related to ozone layer depletion

S o l . Ecological niche was termed by J. Grinnel. It refers the functional role played by the organism where it lives.

116. What type of ecological pyramid would be obtained with the following data?

112. Which of the following is a secondary pollutant?

Secondary consumer : 120 g Primary consumer : 60 g

(1) CO

(2) SO2

Primary producer : 10 g

(3) CO2

(4) O3

(1) Inverted pyramid of biomass

Answer ( 4 )

(2) Upright pyramid of numbers

S o l . O3 (ozone) is a secondary pollutant. These are formed by the reaction of primary pollutant.

(3) Pyramid of energy (4) Upright pyramid of biomass

CO – Quantitative pollutant

Answer ( 1 )

CO2 – Primary pollutant

Sol. •

SO2 – Primary pollutant 113. World Ozone Day is celebrated on

The given data depicts the inverted pyramid of biomass, usually found in aquatic ecosystem.

(1) 5th June

(2) 16th September

•

Pyramid of energy is always upright

(3) 21st April

(4) 22nd April

•

Upright pyramid of biomass and numbers are not possible, as the data depicts primary producer is less than primary consumer and this is less than secondary consumers.

Answer (2) S o l . World Ozone day is celebrated on 16 th September. 5th June - World Environment Day

117. The Golgi complex participates in

21st April - National Yellow Bat Day

(1) Fatty acid breakdown

22nd April - National Earth Day

(2) Respiration in bacteria

114. Natality refers to

(3) Formation of secretory vesicles (4) Activation of amino acid

(1) Death rate (2) Number of individuals leaving the habitat

Answer ( 3 )

(3) Birth rate

S o l . Golgi complex, after processing releases secretory vesicles from their trans-face.

(4) Number of individuals entering a habitat

118. Which of the following is not a product of light reaction of photosynthesis?

Answer ( 3 ) S o l . Natality refers to birth rate.

(1) ATP

(2) NADPH

(3) NADH

(4) Oxygen

•

Death rate

– Mortality

•

Number of individual entering a habitat is

– Immigration

Answer ( 3 )

•

Number of individual leaving the habital

– Emigration

S o l . ATP, NADPH and oxygen are products of light reaction, while NADH is a product of respiration process. 26


119. Which among the following is not a prokaryote? (1) Saccharomyces

(2) Nostoc

(3) Mycobacterium

(4) Oscillatoria

124. Stomata in grass leaf are (1) Dumb-bell shaped (2) Rectangular (3) Kidney shaped

Answer ( 1 )

(4) Barrel shaped

S o l . Saccharomyces i.e. yeast is an eukaryote (unicellular fungi)

Answer ( 1 )

Mycobacterium – a bacterium

S o l . Grass being a monocot, has Dumb-bell shaped stomata in their leaves.

Oscillatoria and Nostoc are cyanobacteria.

125. Which one of the following plants shows a very close relationship with a species of moth, where none of the two can complete its life cycle without the other?

120. Stomatal movement is not affected by (1) Temperature

(2) O2 concentration

(3) Light

(4) CO2 concentration

Answer ( 2 ) S o l . Light, temperature and concentration of CO2 affect opening and closing of stomata while they are not affected by O2 concentration.

(1) Hydrilla

(2) Banana

(3) Yucca

(4) Viola

Answer ( 3 ) S o l . Yucca have an obligate mutualism with a species of moth i.e. Pronuba.

121. Which of the following is true for nucleolus? (1) Larger nucleoli are present in dividing cells

126. Pollen grains can be stored for several years in liquid nitrogen having a temperature of

(2) It takes part in spindle formation (3) It is a membrane-bound structure

(1) –120°C

(2) –196°C

(4) It is a site for active ribosomal RNA synthesis

(3) –80°C

(4) –160°C

Answer ( 2 ) S o l . Pollen grains can be stored for several years in liquid nitrogen at –196°C (Cryopreservation)

Answer ( 4 ) S o l . Nucleolus is a non membranous structure and is a site of r-RNA synthesis.

127. Double fertilization is (1) Fusion of two male gametes of a pollen tube with two different eggs

122. The stage during which separation of the paired homologous chromosomes begins is (1) Pachytene

(2) Diakinesis

(2) Fusion of two male gametes with one egg

(3) Diplotene

(4) Zygotene

(3) Fusion of one male gamete with two polar nuclei

Answer ( 3 )

(4) Syngamy and triple fusion

S o l . Synaptonemal complex disintegrates. Terminalisation begins at diplotene stage i.e. chiasmata start to shift towards end.

Answer ( 4 ) S o l . Double fertilization is a unique phenomenon that occur in angiosperms only.

123. The two functional groups characteristic of sugars are

Syngamy + Triple fusion = Double fertilization 128. Oxygen is not produced during photosynthesis by

(1) Hydroxyl and methyl (2) Carbonyl and phosphate

(1) Green sulphur bacteria

(3) Carbonyl and methyl

(2) Cycas

(4) Carbonyl and hydroxyl

(3) Nostoc

Answer ( 4 )

(4) Chara

S o l . Sugar is a common term used to denote carbohydrate.

Answer ( 1 ) S o l . Green sulphur bacteria do not use H 2O as source of proton, therefore they do not evolve O2.

Carbohydrates are polyhydroxy aldehyde, ketone or their derivatives, which means they have carbonyl and hydroxyl groups. 27


129. Which of the following elements is responsible for maintaining turgor in cells? (1) Magnesium

(2) Potassium

(3) Sodium

(4) Calcium

Sol. 

In Agaricus (a genus of basidiomycetes), basidiospores or meiospores are produced exogenously.



Neurospora (a genus of ascomycetes) produces ascospores as meiospores but endogenously inside the ascus.)



Alternaria (a genus of deuteromycetes) does not produce sexual spores.



Saccharomyces (Unicellular ascomycetes) produces ascospores, endogenously.

Answer ( 2 ) S o l . Potassium helps in maintaining turgidity of cells. 130. What is the role of respiration?

NAD +

in cellular

(1) It functions as an enzyme. (2) It is a nucleotide source for ATP synthesis. (3) It functions as an electron carrier.

134. Which one is wrongly matched?

(4) It is the final electron acceptor for anaerobic respiration.

(1) Uniflagellate gametes – Polysiphonia

Answer ( 3 )

(2) Gemma cups

S o l . In cellular respiration, NAD+ act as an electron carrier.

(3) Biflagellate zoospores – Brown algae (4) Unicellular organism – Chlorella

131. In which of the following forms is iron absorbed by plants?

Answer ( 1 )

(1) Ferric

Sol. •

Polysiphonia is a genus of red algae, where asexual spores and gametes are non-motile or non-flagellated.

•

Other options (2, 3 & 4) are correctly matched

(2) Free element (3) Ferrous (4) Both ferric and ferrous Answer ( 1 * )

135. Match the items given in Column I with those in Column II and select the correct option given below:

S o l . Iron is absorbed by plants in the form of ferric ions. (According to NCERT) *Plants absorb iron in both form i.e. Fe++ and Fe+++. (Preferably Fe++)

Column I

132. Winged pollen grains are present in (1) Mustard

(2) Mango

(3) Cycas

(4) Pinus

– Marchantia

Answer (4)

a. Herbarium

(i) It is a place having a collection of preserved plants and animals

b. Key

(ii) A list that enumerates methodically all the species found in an area with brief description aiding identification

c. Museum

(iii) Is a place where dried and pressed plant specimens mounted on sheets are kept

d. Catalogue

(iv) A booklet containing a list of characters and their alternates which are helpful in identification of various taxa.

S o l . In Pinus, winged pollen grains are present. It is extended outer exine on two lateral sides to form the wings of pollen. It is the characteristic feature, only in Pinus. Pollen grains of Mustard, Cycas & Mango are not winged shaped. 133. After karyogamy followed by meiosis, spores are produced exogenously in (1) Neurospora (2) Agaricus (3) Alternaria (4) Saccharomyces Answer ( 2 ) 28

Column II


a

b

c

d

(1)

(i)

(iv)

(iii)

(ii)

(2)

(ii)

(iv)

(iii)

(i)

(3)

(iii)

(ii)

(i)

(iv)

(4)

(iii)

(iv)

(i)

(ii)

138. The transparent lens in the human eye is held in its place by (1) ligaments attached to the ciliary body (2) smooth muscles attached to the iris (3) ligaments attached to the iris (4) smooth muscles attached to the ciliary body

Answer ( 4 ) Sol. •

Herbarium

–

Dried and pressed plant specimen

•

Key

–

Identification of various taxa

S o l . Lens in the human eye is held in its place by suspensory ligaments attached to the ciliary body.

•

Museum

–

Plant and animal specimen are preserved

139. Which of the following hormones can play a significant role in osteoporosis?

•

Catalogue

–

Alphabetical listing of species

Answer ( 1 )

(1) Aldosterone and Prolactin (2) Estrogen and Parathyroid hormone (3) Progesterone and Aldosterone

136. Which of the following is an amino acid derived hormone?

(4) Parathyroid hormone and Prolactin

(1) Epinephrine

(2) Estradiol

Answer ( 2 )

(3) Ecdysone

(4) Estriol

S o l . Estrogen promotes the activity of osteoblast and inhibits osteoclast. In an ageing female osteoporosis occurs due to deficiency of estrogen. Parathormone promotes mobilisation of calcium from bone into blood. Excessive activity of parathormone causes demineralisation leading to osteoporosis.

Answer ( 1 ) S o l . Epinephrine is derived from tyrosine amino acid by the removal of carboxyl group. It is a catecholamine. 137. Which of the following structures or regions is incorrectly paired with its functions?

140. Among the following sets of examples for divergent evolution, select the incorrect option :

(1) Medulla oblongata : controls respiration and cardiovascular reflexes. (2) Hypothalamus

(1) Forelimbs of man, bat and cheetah (2) Brain of bat, man and cheetah

: production of releasing hormones and regulation of temperature, hunger and thirst.

(3) Limbic system

: consists of fibre tracts that interconnect different regions of brain; controls movement.

(4) Corpus callosum

: band of fibers connecting left and right cerebral hemispheres.

(3) Heart of bat, man and cheetah (4) Eye of octopus, bat and man Answer ( 4 ) S o l . Divergent evolution occurs in the same structure, example - forelimbs, heart, brain of vertebrates which have developed along different directions due to adaptation to different needs whereas eye of octopus, bat and man are examples of analogous organs showing convergent evolution. 141. Which of the following is not an autoimmune disease? (1) Psoriasis (2) Alzheimer's disease (3) Rheumatoid arthritis

Answer ( 3 )

(4) Vitiligo

S o l . Limbic system is emotional brain. It controls all emotions in our body but not movements.

Answer ( 2 ) 29


S o l . Rheumatoid arthritis is an autoimmune disorder in which antibodies are produced against the synovial membrane and cartilage.

145. Conversion of milk to curd improves its nutritional value by increasing the amount of

Vitiligo causes white patches on skin also characterised as autoimmune disorder.

(1) Vitamin D

(2) Vitamin B12

(3) Vitamin A

(4) Vitamin E

Answer ( 2 )

Psoriasis is a skin disease that causes itchy or sore patches of thick red skin and is also autoimmune whereas Alzheimer's disease is due to deficiency of neurotransmitter acetylcholine.

Sol.  

Curd is more nourishing than milk. It has enriched presence of vitamins specially Vit-B12.

146. Which one of the following population interactions is widely used in medical science for the production of antibiotics?

142. Which of the following characteristics represent ‘Inheritance of blood groups’ in humans? a. Dominance

(1) Commensalism

(2) Parasitism

(3) Mutualism

(4) Amensalism

b. Co-dominance

Answer ( 4 )

c. Multiple allele

S o l . Amensalism/Antibiosis (0, –)

d. Incomplete dominance



Antibiotics are chemicals secreted by one microbial group (eg : Penicillium) which harm other microbes (eg : Staphylococcus)



It has no effect on Penicillium or the organism which produces it.

e. Polygenic inheritance (1) b, c and e

(2) b, d and e

(3) a, b and c

(4) a, c and e

Answer ( 3 ) Sol. 

IAIO, IBIO

-

Dominant–recessive relationship



IAIB

-

Codominance

(1) Wildlife safari parks



IA, IB

-

3-different allelic forms of a gene (multiple allelism)

(2) Botanical gardens

&

IO

147. All of the following are included in ‘ex-situ conservation’ except

(3) Sacred groves (4) Seed banks

143. In which disease does mosquito transmitted pathogen cause chronic inflammation of lymphatic vessels? (1) Elephantiasis

(2) Ringworm disease

(3) Ascariasis

(4) Amoebiasis

Answer ( 3 ) Sol.  

Sacred groves – in-situ conservation. Represent pristine forest patch as protected by Tribal groups.

148. Match the items given in Column I with those in Column II and select the correct option given below :

Answer ( 1 ) S o l . Elephantiasis is caused by roundworm, Wuchereria bancrofti and it is transmitted by Culex mosquito.

Column-I

144. The similarity of bone structure in the forelimbs of many vertebrates is an example of

Column-II

a. Eutrophication

i.

b. Sanitary landfill

ii. Deforestation

c. Snow blindness

iii. Nutrient enrichment

(1) Homology (2) Convergent evolution

UV-B radiation

d. Jhum cultivation iv. Waste disposal

(3) Analogy

a

(4) Adaptive radiation

b

c

d

(1) ii

i

iii

iv

Answer ( 1 )

(2) iii

iv

i

ii

S o l . In different vertebrates, bones of forelimbs are similar but their forelimbs are adapted in different way as per their adaptation, hence example of homology.

(3) i

iii

iv

ii

(4) i

ii

iv

iii

Answer ( 2 ) 30


S o l . a. Eutrophication

iii.

Nutrient enrichment

b. Sanitary landfill

iv.

Waste disposal

(1) blocks estrogen receptors in the uterus, preventing eggs from getting implanted.

c. Snow blindness

i.

UV-B radiation

(2) is an IUD.

Deforestation

(3) increases the concentration of estrogen and prevents ovulation in females.

d. Jhum cultivation ii.

152. The contraceptive ‘SAHELI’

149. In a growing population of a country, (1) pre-reproductive individuals are more than the reproductive individuals.

(4) is a post-coital contraceptive. Answer ( 1 )

(2) reproductive and pre-reproductive individuals are equal in number.

S o l . Saheli is the first non-steroidal, once a week pill. It contains centchroman and its functioning is based upon selective Estrogen Receptor modulation.

(3) reproductive individuals are less than the post-reproductive individuals. (4) pre-reproductive individuals are less than the reproductive individuals.

153. The amnion of mammalian embryo is derived from

Answer ( 1 )

(1) ectoderm and mesoderm

S o l . Whenever the pre-reproductive individuals or the younger population size is larger than the reproductive group, the population will be an increasing population.

(2) mesoderm and trophoblast (3) endoderm and mesoderm (4) ectoderm and endoderm

150. Which part of poppy plant is used to obtain the drug “Smack”? (1) Flowers

(2) Roots

(3) Latex

(4) Leaves

Answer ( 1 ) S o l . The extraembryonic or foetal membranes are amnion, chorion, allantois and Yolk sac. Amnion is formed from mesoderm on outer side and ectoderm on inner side.

Answer ( 3 ) S o l . ‘Smack’ also called as brown sugar/Heroin is formed by acetylation of morphine. It is obtained from the latex of unripe capsule of Poppy plant.

Chorion is formed from trophoectoderm and mesoderm whereas allantois and Yolk sac membrane have mesoderm on outerside and endoderm in inner side.

151. Hormones secreted by the placenta to maintain pregnancy are

154. The difference between spermiogenesis and spermiation is

(1) hCG, hPL, progestogens, prolactin

(1) In spermiogenesis spermatids are formed, while in spermiation spermatozoa are formed.

(2) hCG, hPL, progestogens, estrogens (3) hCG, hPL, estrogens, relaxin, oxytocin (4) hCG, progestogens, glucocorticoids

estrogens,

(2) In spermiogenesis spermatozoa from sertoli cells are released into the cavity of seminiferous tubules, while in spermiation spermatozoa are formed.

Answer ( 2 ) S o l . Placenta releases human chorionic gonadotropic hormone (hCG) which stimulates the Corpus luteum during pregnancy to release estrogen and progesterone and also rescues corpus luteum from regression. Human placental lactogen (hPL) is involved in growth of body of mother and breast. Progesterone maintains pregnancy, keeps the uterus silent by increasing uterine threshold to contractile stimuli.

(3) In spermiogenesis spermatozoa are formed, while in spermiation spermatids are formed. (4) In spermiogenesis spermatozoa are formed, while in spermiation spermatozoa are released from sertoli cells into the cavity of seminiferous tubules. Answer ( 4 ) 31


S o l . Spermiogenesis is transformation of spermatids into spermatozoa whereas spermiation is the release of the sperms from sertoli cells into the lumen of seminiferous tubule.

157. Match the items given in Column I with those in Column II and select the correct option given below: Column I

155. Which of the following options correctly represents the lung conditions in asthma and emphysema, respectively?

a

surface;

a. Tricuspid valve

i.

b. Bicuspid valve

ii. Between right ventricle and pulmonary artery

c. Semilunar valve

iii. Between right atrium and right ventricle

(1) iii

i

ii

(2) i

ii

iii

(3)

iii

ii

i

iii

i

(4) ii

iv. 1000 – 1100 mL

b

c

d

(1) iii

ii

i

iv

(2) i

iv

ii

iii

(3) iii

i

iv

ii

(4) iv

iii

ii

i

Answer ( 3 ) S o l . Tidal volume is volume of air inspired or expired during normal respiration. It is approximately 500 mL. Inspiratory reserve volume is additional volume of air a person can inspire by a forceful inspiration. It is around 2500 – 3000 mL. Expiratory reserve volume is additional volume of air a person can be expired by a forceful expiration. This averages 1000 – 1100 mL.

Column II

c

iii. 500 – 550 mL

d. Residual volume


b

ii. 1100 – 1200 mL

volume

S o l . Asthma is a difficulty in breathing causing wheezing due to inflammation of bronchi and bronchioles. Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased.

a

b. Inspiratory Reserve

c. Expiratory Reserve

Answer ( 1 )

Column I

i. 2500 – 3000 mL

surface;

(3) Increased number of bronchioles; Increased respiratory surface (4) Decreased respiratory Inflammation of bronchioles

a. Tidal volume

volume

(1) Inflammation of bronchioles; Decreased respiratory surface (2) Increased respiratory Inflammation of bronchioles

Column II

Between left atrium and left ventricle

Residual volume is volume of air remaining in lungs even after forceful expiration. This averages 1100 – 1200 mL. 158. Match the items given in Column I with those in Column II and select the correct option given below : Column I

Answer ( 1 ) S o l . Tricuspid valves are AV valve present between right atrium and right ventricle. Bicuspid valves are AV valve present between left atrium and left ventricle. Semilunar valves are present at the openings of aortic and pulmonary aorta. 32

Column II

a. Glycosuria

i.

Accumulation of uric acid in joints

b. Gout

ii. Mass of crystallised salts within the kidney

c. Renal calculi

iii. Inflammation in glomeruli

d. Glomerular nephritis

iv. Presence of in glucose urine


a

b

c

d

(1)

iii

ii

iv

i

(2)

ii

iii

i

iv

(3)

i

ii

iii

iv

(4)

iv

i

ii

iii

Urine is carried from kidney to bladder through ureter. Urinary bladder is concerned with storage of urine. 160. Which of the following events does not occur in rough endoplasmic reticulum?

Answer ( 4 )

(1) Protein folding

S o l . Glycosuria denotes presence of glucose in the urine. This is observed when blood glucose level rises above 180 mg/100 ml of blood, this is called renal threshold value for glucose. Gout is due to deposition of uric acid crystals in the joint.

(2) Cleavage of signal peptide (3) Protein glycosylation (4) Phospholipid synthesis Answer ( 4 ) S o l . Phospholipid synthesis does not take place in RER. Smooth endoplasmic reticulum are involved in lipid synthesis.

Renal calculi are precipitates of calcium phosphate produced in the pelvis of the kidney. Glomerular nephritis is the inflammatory condition of glomerulus characterised by proteinuria and haematuria.

161. Which of these statements is incorrect? (1) Enzymes of TCA cycle are present in mitochondrial matrix

159. Match the items given in Column I with those in Column II and select the correct option given below: Column I

Column II

(Function)

(Part of Excretory system)

a. Ultrafiltration

i.

b. Concentration

ii. Ureter

(2) Glycolysis operates as long as it is supplied with NAD that can pick up hydrogen atoms (3) Glycolysis occurs in cytosol (4) Oxidative phosphorylation takes place in outer mitochondrial membrane

Henle's loop

Answer ( 4 ) S o l . Oxidative phosphorylation takes place in inner mitochondrial membrane.

of urine c. Transport of

iii. Urinary bladder

162. Nissl bodies are mainly composed of

urine

(1) Proteins and lipids

d. Storage of

iv. Malpighian

urine

(2) Nucleic acids and SER

corpuscle

(3) DNA and RNA

v. Proximal

(4) Free ribosomes and RER

convoluted tubule a

b

c

d

(1) iv

v

ii

iii

(2) v

iv

i

ii

(3) iv

i

ii

iii

(4) v

iv

i

iii

Answer ( 4 ) S o l . Nissl granules are present in the cyton and even extend into the dendrite but absent in axon and rest of the neuron. Nissl granules are in fact composed of free ribosomes and RER. They are responsible for protein synthesis.

Answer ( 3 )

163. Which of the following terms describe human dentition?

S o l . Ultrafiltration refers to filtration of very fine particles having molecular weight less than 68,000 daltons through malpighian corpuscle.

(1) Thecodont, Diphyodont, Homodont (2) Pleurodont, Monophyodont, Homodont

Concentration of urine refers to water absorption from glomerular filtrate as a result of hyperosmolarity in the medulla created by counter-current mechanism in Henle's loop.

(3) Thecodont, Diphyodont, Heterodont (4) Pleurodont, Diphyodont, Heterodont Answer ( 3 ) 33



S o l . In humans, dentition is 

Thecodont : Teeth are present in the sockets of the jaw bone called alveoli.



Diphyodont : Teeth erupts twice, temporary milk or deciduous teeth are replaced by a set of permanent or adult teeth.



Column I

endometrial lining

Heterodont dentition : Dentition consists of different types of teeth namely incisors, canine, premolars and molars.

b. Secretory Phase

ii. Follicular Phase

c. Menstruation

iii. Luteal Phase

a

164. Select the incorrect match : (1) Lampbrush

Column II

a. Proliferative Phase i. Breakdown of

– Diplotene bivalents

chromosomes

b

c

(1) iii

ii

i

(2) ii

iii

i

(3) i

iii

ii

(4) iii

i

ii

(2) Submetacentric – L-shaped chromosomes chromosomes

Answer ( 2 )

(3) Allosomes

– Sex chromosomes

(4) Polytene chromosomes

– Oocytes of amphibians

S o l . During proliferative phase, the follicles start developing, hence, called follicular phase. Secretory phase is also called as luteal phase mainly controlled by progesterone secreted by corpus luteum. Estrogen further thickens the endometrium maintained by progesterone.

Answer ( 4 ) S o l . Polytene chromosomes are found in salivary glands of insects of order Diptera.

Menstruation occurs due to decline in progesterone level and involves breakdown of overgrown endometrial lining.

165. Many ribosomes may associate with a single mRNA to form multiple copies of a polypeptide simultaneously. Such strings of ribosomes are termed as

168. All of the following are part of an operon except

(1) Polysome

(1) an operator

(2) Plastidome

(2) an enhancer

(3) Polyhedral bodies

(3) structural genes

(4) Nucleosome

(4) a promoter

Answer ( 1 )

Answer ( 2 )

S o l . The phenomenon of association of many ribosomes with single m-RNA leads to formation of polyribosomes or polysomes or ergasomes.

Sol. • •

Enhancer sequences are present in eukaryotes. Operon concept is for prokaryotes.

169. AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequence of the transcribed mRNA?

166. According to Hugo de Vries, the mechanism of evolution is (1) Multiple step mutations (2) Phenotypic variations

(1) AGGUAUCGCAU

(3) Saltation

(2) ACCUAUGCGAU

(4) Minor mutations

(3) UGGTUTCGCAT (4) UCCAUAGCGUA

Answer ( 3 )

Answer ( 1 )

S o l . As per mutation theory given by Hugo de Vries, the evolution is a discontinuous phenomenon or saltatory phenomenon/ saltation.

S o l . Coding strand and mRNA has same nucleotide sequence except, ‘T’ – Thymine is replaced by ‘U’–Uracil in mRNA. 34


Antibodies are derived from -Globulin fraction of plasma proteins which means globulins are involved in defence mechanisms.

170. A woman has an X-linked condition on one of her X chromosomes. This chromosome can be inherited by (1) Only daughters

Albumin is a plasma responsible for BCOP.

(2) Only grandchildren (3) Only sons

protein

173. Which of the following is an occupational respiratory disorder?

(4) Both sons and daughters Answer ( 4 )

(1) Anthracis

(2) Botulism

Sol. •

(3) Silicosis

(4) Emphysema

Woman is a carrier

•

Both son X–chromosome

&

•

Although only son be the diseased

daughter inherit

Answer ( 3 ) S o l . Silicosis is due to excess inhalation of silica dust in the workers involved grinding or stone breaking industries.

171. Which of the following gastric cells indirectly help in erythropoiesis? (1) Chief cells

(2) Goblet cells

(3) Mucous cells

(4) Parietal cells

Long exposure can give rise to inflammation leading to fibrosis and thus causing serious lung damage. Anthrax is a serious infectious disease caused by Bacillus anthracis. It commonly affects domestic and wild animals. Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased.

Answer ( 4 ) S o l . Parietal or oxyntic cell is a source of HCl and intrinsic factor. HCl converts iron present in diet from ferric to ferrous form so that it can be absorbed easily and used during erythropoiesis.

Botulism is a form of food poisoning caused by Clostridium botulinum.

Intrinsic factor is essential for the absorption of vitamin B 12 and its deficiency causes pernicious anaemia.

174. Calcium is important in skeletal muscle contraction because it

172. Match the items given in Column I with those in Column II and select the correct option given below : Column I

(i) Osmotic balance

b. Globulin

(ii) Blood clotting

c. Albumin

(iii) Defence mechanism

a

(1) Binds to troponin to remove the masking of active sites on actin for myosin. (2) Detaches the myosin head from the actin filament.

Column II

a. Fibrinogen

b

c

(1) (iii)

(ii)

(i)

(2) (i)

(iii)

(ii)

(3) (i)

(ii)

(iii)

(4) (ii)

(iii)

(i)

mainly

(3) Activates the myosin ATPase by binding to it. (4) Prevents the formation of bonds between the myosin cross bridges and the actin filament. Answer ( 1 ) Sol. 

Signal for contraction increase Ca++ level many folds in the sarcoplasm.



Ca++ now binds with sub-unit of troponin (troponin "C") which is masking the active site on actin filament and displaces the sub-unit of troponin.



Once the active site is exposed, head of the myosin attaches and initiate contraction by sliding the actin over myosin.

Answer ( 4 ) S o l . Fibrinogen forms fibrin strands during coagulation. These strands forms a network and the meshes of which are occupied by blood cells, this structure finally forms a clot. 35


175. Identify the vertebrate group of animals characterized by crop and gizzard in its digestive system (1) Amphibia

(2) Aves

(3) Reptilia

(4) Osteichthyes

178. Which one of these animals is not a homeotherm? (1) Macropus (2) Camelus (3) Chelone

Answer ( 2 )

(4) Psittacula

S o l . The digestive tract of Aves has additional chambers in their digestive system as crop and Gizzard.

Answer ( 3 ) S o l . Homeotherm are animals that maintain constant body temperature, irrespective of surrounding temperature.

Crop is concerned with storage of food grains. Gizzard is a masticatory organ in birds used to crush food grain.

Birds and mammals are homeotherm. Chelone (Turtle) belongs to class reptilia which is Poikilotherm or cold blood.

176. Ciliates differ from all other protozoans in (1) using flagella for locomotion

179. Which of the following animals does not undergo metamorphosis?

(2) using pseudopodia for capturing prey (3) having a contractile vacuole for removing excess water

(1) Earthworm (2) Moth

(4) having two types of nuclei

(3) Tunicate

Answer ( 4 )

(4) Starfish

S o l . Ciliates differs from other protozoans in having two types of nuclei.

Answer ( 1 ) S o l . Metamorphosis refers to transformation of larva into adult.

eg. Paramoecium have two types of nuclei i.e. macronucleus & micronucleus.

Animal that perform metamorphosis are said to have indirect development.

177. Which of the following features is used to identify a male cockroach from a female cockroach?

In earthworm development is direct which means no larval stage and hence no metamorphosis.

(1) Presence of a boat shaped sternum on the 9th abdominal segment

180. Which of the following organisms are known as chief producers in the oceans?

(2) Forewings with darker tegmina (3) Presence of caudal styles

(1) Dinoflagellates

(4) Presence of anal cerci

(2) Cyanobacteria

Answer ( 3 )

(3) Diatoms

S o l . Males bear a pair of short, thread like anal styles which are absent in females.

(4) Euglenoids Answer ( 3 )

Anal/caudal styles arise from 9th abdominal segment in male cockroach.

S o l . Diatoms are chief producers of the ocean.

36